144.44 in² is the closest amount of paper she will use to wrap the gift.
Solution:
The given net is the shape of the cylinder.
Diameter of the circle = 4 in
Radius of the circle = 4 ÷ 2 = 2 in
Width of the container = 10.5 in
Length of the container is same as the circumference of the circle.
Area of the circle = 
= 
= 12.56 in²
Area of the 2 circles = 2 × 12.56 = 25.12 in²
Area of the rectangle = circumference × height
= 
= 
= 131.88 in²
Area of the net of the container
= Area of the circles + Area of the rectangle
= 12.56 in² + 131.88 in²
= 144.44 in²
Area of the net of the container = 144.44 in²
Hence 144.44 in² is the closest amount of paper she will use to wrap the gift.
Answer:
56
Step-by-step explanation:
Missing length = x + 21
Based on the Three Parallel Lines Theorem, we have:
x/21 = 20/12
Cross multiply
x*12 = 20*21
12x = 420
12x/12 = 420/12
x = 35
✅Missing length = x + 21
Plug in the value of x
= 35 + 21
= 56
The answer is 0.45 recurring. Hope this helped or helps you. :)
Answer:
24 cm²
Step-by-step explanation:
The rectangular section on the right is 8 - 4 = 4 cm wide.
It is 3 + 1.5 = 4.5 cm tall.
The area of the rectangular section is ...
A = LW
A = (4.5 cm)(4 cm) = 18 cm²
__
The triangular section on the left is 4 cm wide and 3 cm high. Its area is ...
A = 1/2bh
A = 1/2(4 cm)(3 cm) = 6 cm²
The total shaded area is the sum of the areas of the triangular section and the rectangular section:
area = 18 cm² +6 cm² = 24 cm² . . . area of the figure